Suppose that $f$ and $g$ are two commuting continuous mappings from the closed unit disk (or, if you prefer, the closed unit ball in $R^n$) to itself. Does there always exist a point $x$ such that $f(x)=g(x)$?

If one of the mappings is invertible, then it is just a restatement of Brouwer's fixed point theorem but I do not know the answer in the general case and would not even dare to guess what it must be. Also, the answer is well-known to be "Yes" in dimension $1$.

Could more be true? Would f and g necessarily have a common fixed point? This might perhaps be easier to prove if true.
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Harald Hanche-OlsenOct 29 '09 at 23:35

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Apparently f and g may not have a common fixed point even in the dimension 1 case. This is mentioned in the first paragraph of "Equivalent Conditions involving Common Fixed Points in the Unit Interval" by Jachymski. Unfortunately I can't follow the reference given. @fedja, what an amazing problem!
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Alon AmitOct 30 '09 at 16:58

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Ah, indeed: Jachymski refers to an abstract in an old Notices. But searching MR reveals J.P. Huneke: On common fixed points of commuting continuous functions on an interval. Trans. Amer. Math. Soc. 139 1969 371--381. See jstor.org/stable/1995330 if you have JSTOR access. From the abstract: “This paper offers two methods of constructing commuting pairs of continuous functions [...] which map [0,1] to itself without common fixed points”. Jachymski also notes that if the iterates of one function forms an equicontinuous family, there is a common fixed point.
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Harald Hanche-OlsenOct 30 '09 at 18:02

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If I understand well, you don't ask whether $f$ and $g$ have a common fixed point. Yet, the answers given so far speak of fixed points ...
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Denis SerreMar 31 '11 at 14:28

8 Answers
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I'm sure what I write has been thought of by many, but it's a starting point that I thought should be written down.

First by the Brouwer fixed point theorem $f$ has at least one fixed point, say $\bar{x}=f(\bar{x})$.

If that fixed point is unique (contraction mappings spring to mind for a bunch of examples of this) we're done since $g(\bar{x})=g(f(\bar{x}))=f(g(\bar{x}))$ and we see that $g(\bar{x})$ is "another" fixed point of $f$, since the fixed point was unique $g(\bar{x})=\bar{x}=f(\bar{x})$.

For "less nice" $f$ we still have that $f(g(\bar{x}))=g(\bar{x})$... in fact for any $n\in\mathbb{N}$ we have $f(g^n(\bar{x}))=g^n(\bar{x})$. If (without resorting to sequences) $g^n(\bar{x})\to y$, we can again claim success since we'll have $g(y)=y$ and $f(y)=y$.

Unless there's another "obvious" easy case I missed it seems like the interesting cases will be when $g^n(\bar{x})$ does not converge. Two sub-cases spring to mind: when $g^{n}(\bar{x})$ has finitely many accumulation points (like when $g^n(\bar{x})$ is a periodic point of $g$), or ... it has lots. Intuition (really thinking about rational and then irrational rotations about the origin as one way to generate those two cases) tells me that in either of these cases what we really need to do drop the $\bar{x}$ as a "starting point".

It "would be nice" if we can show $g$ conjugate to a rotation in the above two cases. Any thought on if that is true or not? I suspect not else $g$ would have a unique fixed point and we'd be done (as above)... Maybe semi-conjugate... but would that help? New minds, any thoughts?

$g$ could be a reflection instead of a rotation. Draw a vector field in the disc which is symmetric across the $y$-axis and which has exactly two zeros, and let $f$ be a small flow along this vector field. Let $g$ be reflection in the $y$-axis. There are exactly two places $f$ and $g$ agree: $(0,1)$ and $(0,-1)$.
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Sean EberhardApr 13 at 16:16

A positive answer to this question would imply a positive answer to the open problem on existence of a "coincidence point" of two commuting maps $f_1, f_2: T\to T$, where $T$ is the triod (tripod), see Question 1 in McDowell's survey http://topology.auburn.edu/tp/reprints/v34/tp34025p1.pdf

The point is that if $f_1, f_2: T\to T$ are commuting maps as above, one can define maps $\tilde{f_i}=f_i\circ R: D^2\to D^2$ where $R: D^2\to T$ is a retraction. (I am assuming that $T$ is embedded in $D^2$.) Then $\tilde{f_1}, \tilde{f_2}$ commute if and only if $f_1$ and $f_2$ commute, furthermore, $f_1$ and $f_2$ have a coincidence point $x, f_1(x)=f_2(x),$ if and only if $\tilde{f_1}, \tilde{f_2}$ do.

Of course, if one were to look for counter-examples, it would be easier to find ones among maps of the 2-disk.

The following is not a solution but rather a reformulation of the original problem.

To make $f\colon B\to B$ invertible let us pass from the ball $B$ to the solenoid $S_f$,
$$
S_f=\{\{x_n; n\in\mathbb Z\}: x_{n+1}=f(x_n)\}
$$

Map $f$ induces the shift $f_*\colon S_f\to S_f$ and g induces $g_*(\{x_n\})=\{g(x_n)\}$. They commute and $f_*$ is invertible. If Brower's fixed point theorem
were true for $S_f$ the result would follow. Indeed, a fixed point of $f_*^{-1}\circ g_*$ gives the desired orbit.

Solenoid $S_f$ is a non-empty closed subset of $B^{\mathbb Z}$ eqiupped with the product topology. Hence $S_f$ is compact. In fact, $S_f\subset K^{\mathbb Z}$,
where $K=\cap_{n>0}f^n(B)$. And I think it is plausible (can anybody give a proof?) that $S_f$ contracts to the orbit of the fixed point of $f$. (later: I am not that sure now)

Could you put that pdf somewhere for those of us who aren't at utexas? Well, or at least the title of the paper so we can find it through our own institutions? At the moment, that link only works for utexas people.
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Harry GindiJun 13 '10 at 8:59

Well, I give up trying to fix the formula at the end of the first paragraph.
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Andrey GogolevJun 15 '10 at 17:47

May be a possible way to solve the problem is to set up it in the context of closed relations instead of functions, by putting $R = g^{-1} \circ f : B^n \to B^n$, and assuming for instance $f(B^n) \subset g(B^n)$. So, we are looking for $x \in B^n$ such that $x \in R(x)$.

Under nice circumstances the multihomotopy groups of spheres are $\tilde\pi_n(S^n) \cong \Bbb Z$, while for the $n$-disk are trivial.

One can try to adapt the proof of Brouwer's fixed point theorem in this setting.

So, assuming by contradicition that $f(x) \ne g(x)$ for all $x \in B^n$, one can consider the relation $T : B^n \to S^{n-1}$ where $T(x)$ is the intersection between $S^{n-1}$ and the half-lines starting from points of $R(x)$, and passing through $x$. It follows that $T$ is the identity on $S^{n-1}$, so $T$ is a multi-valued retraction $B^n \to S^{n-1}$, and this contradicts the fact that $\tilde\pi_{n-1}(S^{n-1})$ is not trivial.

I think that there are many details to fill, and that one should give the "nice" conditions under this approach works (for instance, assuming that $g^{-1}(y)$ is countable for all $y \in B^n$ does suffice?) but this could be an interesting application of multihomotopy groups.

I made a little google search and found this paper, that mentions that the conjecture holds true for polynomials and special functions Cohen calls "full functions". I'm not able to download the paper though. Cohen's paper is: H. Cohen, On fixed points of commuting functions, Proc. Amer. Math. Soc. 15. (1964),

An idea that occured to me, since apparently this is true for polynomials. Can we not use some form of approximation theorem (Stone-Weierstrass or whatever there is) to conclude the result for other continuous functions between the closed unit balls?

By the Brouwer fixed point theorem, for every f there is an x such that f(x) = x, so the answer is yes when g is the identity.
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Richard KentNov 6 '09 at 12:58

Ok sorry.. Kent is correct.. Brouwers fixed point theorem, my argument does not work because the function that I "drew" below the diagonal must have values in -0.5 an 0.5 respectively so it will necessarily cross the diagonal..
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Jose CapcoNov 6 '09 at 13:48

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The trouble with proving this via an approximation scheme is that you would have to prove that if $f$ and $g$ commute, then you can approximate $f$ and $g$ by polynomials $f'$ and $g'$ that commute. This sounds unlikely.
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Andy PutmanNov 6 '09 at 18:10

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I think that you should avoid editing your answers the way you did (erasing everything and writing a new answer). It's better to delete the answer and write a new one.
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Ori Gurel-GurevichNov 7 '09 at 7:47

I wasn't able to delete after several posts.. and things get being added to the features I am able to do here. I think this is a *overlflow.net system, it's a bit confusing and annoying. I wasn't even able to make comments before. But now I can at least delete :)
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Jose CapcoNov 7 '09 at 9:34